# Matrices proof

• Jan 18th 2013, 07:25 PM
Tutu
Matrices proof
If X = \$\displaystyle P^{-1}AP and A^{3} = I, prove that X^{3} =I\$

I did this
\$\displaystyle X^{3} = (P^{-1}AP)(P^{-1}AP)(P^{-1}AP) = (P^{-3}A^{3}P^{3}) = (P^{-3+3}A^{3}) = A^{3} = I\$

I used the law of indices, is it right to apply this for matrices?

Thank you!
• Jan 18th 2013, 07:55 PM
chiro
Re: Matrices proof
Hey Tutu.

Hint: Group the P*P^(-1) = I (I is identity matrix) together to get a lot of cancellation and then use A^3 = I to get your final result.
• Jan 21st 2013, 03:53 AM
Tutu
Re: Matrices proof
Thank you!
I thought I could not change the order within the multiplication, but so is it that I can do it in thie case? Why?
Is it
X^3 = (P^(-1)PA)(P^(-1)PA)(P^(-1)PA)
= P^(-3) (PA)^3
= I
Wait, I did this using laws of indices again, can you please explain how to do it, I'm afraid I do not understand..

Thank you ever so much!
• Jan 21st 2013, 06:08 AM
Deveno
Re: Matrices proof

\$\displaystyle X^3 = (P^{-1}AP)^3 = (P^{-1}AP)(P^{-1}AP)(P^{-1}AP)\$

\$\displaystyle = P^{-1}A(PP^{-1})A(PP^{-1})AP = P^{-1}AIAIAP = P^{-1}AAAP = P^{-1}A^3P\$

NOT \$\displaystyle P^{-3}A^3P^3\$.

and since \$\displaystyle A^3 = I\$,

\$\displaystyle P^{-1}A^3P = P^{-1}IP = P^{-1}P = I\$.

no "switching around of order" is required.
• Jan 21st 2013, 06:56 AM
Tutu
Re: Matrices proof
Thank you!
Does that mean that in matrices, PP^(-1) = P? I thought it equalled to I
• Jan 21st 2013, 06:57 AM
Tutu
Re: Matrices proof
Sorry, ignore the previous post. I get it now!